POJ 1985-Cow Marathon【树的直径模板】

分手后的思念是犯贱 2022-08-21 10:39 13阅读 0赞

Cow Marathon

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;2000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;30000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;4583</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;2265</td> 
  </tr> 
  <tr> 
   <td colspan="3" align="center" style="color:rgb(14,173,0)"><strong>Case Time Limit:</strong>&nbsp;1000MS</td> 
  </tr> 
 </tbody> 
</table>

Description

After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has committed to create a bovine marathon for his cows to run. The marathon route will include a pair of farms and a path comprised of a sequence of roads between them. Since FJ wants the cows to get as much exercise as possible he wants to find the two farms on his map that are the farthest apart from each other (distance being measured in terms of total length of road on the path between the two farms). Help him determine the distances between this farthest pair of farms.

Input

\* Lines 1.....: Same input format as "Navigation Nightmare".

Output

\* Line 1: An integer giving the distance between the farthest pair of farms.

Sample Input

7 6
    1 6 13 E
    6 3 9 E
    3 5 7 S
    4 1 3 N
    2 4 20 W
    4 7 2 S

Sample Output

Hint

The longest marathon runs from farm 2 via roads 4, 1, 6 and 3 to farm 5 and is of length 20+3+13+9+7=52.

解题思路：

树的直径，来，上模板。不用考虑最后那个方向，就是拿来逗人的。

#include <iostream>
    #include<stdio.h>
    #include <string.h>
    #define MAXN 60005
    #define INF 0x3f3f3f3f
    
    using namespace std;
    
    struct Edge
    {
        int v, w, next;
    }edge[MAXN*6];
    
    int m,head[MAXN], vis[MAXN], d[MAXN], q[MAXN], e, n;
    
    int init()
    {
        e = 0;
        memset(head, -1, sizeof(head));
    }
    
    void add(int u, int v, int w)
    {
        edge[e].v = v;
        edge[e].w = w;
        edge[e].next = head[u];
        head[u] = e++;
    }
    
    void bfs(int src)
    {
        for (int i = 1; i <= n; i++)
        {
            vis[i] = 0;
            d[i] = INF;
        }
    
        int h = 0, t = 0;
        vis[src] = 1;
        q[t++] = src;
        d[src] = 0;
        while (h < t)
        {
            int u = q[h++];
            for (int i = head[u]; i != -1; i = edge[i].next)
            {
                int v = edge[i].v;
                int w = edge[i].w;
                if (d[u] + w < d[v])
                {
                    d[v] = d[u] + w;
                    if (!vis[v])
                    {
                        q[t++] = v;
                        vis[v] = 1;
                    }
                }
            }
        }
    }
    
    int main()
    {
        int u, v, w;
    	while(scanf("%d%d",&n,&m)!=EOF)
    	{
    		init();
    		char pp;
    	    for (int i=1;i<=m; i++)
    	    {
    	        scanf("%d%d%d %c",&u,&v,&w,&pp);
    	        add(u, v, w);
    	        add(v, u, w);
    	    }
    	    bfs(1);
    	    int pos = 0, mx = -1;
    	    for (int i  = 1; i <= n; i++)
    	    {
    	        if(d[i] > mx)
    	        {
    	            mx = d[i];
    	            pos = i;
    	        }
    	    }
    	    bfs(pos);
    	    mx = -1;
    	    for (int i  = 1; i <= n; i++)
    	    {
    	        if(d[i] > mx)
    	        {
    	            mx = d[i];
    	        }
    	    }
    	    printf("%d\n",mx);
    	} 
        return 0;
    }